Integration by substitution (and esp. Weierstrass' substitution)

  • Thread starter dodo
  • Start date
695
2
The stupid question of the day.

Is it fair to say that[tex]\frac{du}{dx} = \frac 1 {dx/du}
[/tex]since this comes (I think) from the chain rule,[tex]
\frac{dx}{du} \frac{du}{dx} = \frac{dx}{dx} = 1
[/tex]

Which means that, when integrating by substitution, I can choose to do either of[tex]
\int f(u) du = \int f(x) \frac {du}{dx} dx = \int \frac {f(x)} {dx/du} dx
[/tex]depending on which derivative I happen to have at hand.

(Just checking; you can't be too careful when treating differentials as it they were fractions, which they aren't.)

The matter came out while studying Weierstrass' substitution, where[tex]
\begin{align*}
\int \frac 1 {\sin x} dx &= \int \frac {\frac{1+t^2}{2t}} {\frac{1+t^2}{2}}
= \int \frac 1 t dt &\mbox{with }t=\tan \frac x 2
\end{align*}
[/tex]because [itex]\frac {dt}{dx}[/itex] is easier to figure out than [itex]\frac {dx}{dt}[/itex].

Any more detail (particularly from the viewpoint of analysis) is welcome. Right now I feel like I'm just shuffling symbols without really knowing what am I doing.
 

HallsofIvy

Science Advisor
Homework Helper
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As long as u(x) is invertible (i.e. as long as du/dx is not 0), yes, [itex]dx/du= 1/(du/dx)[/itex].

Your second statement is also correct but you have to be careful with it. Unless du/dx is a constant we cannot just shift du/dx inside or outside the integral. But we can use one or the other of those forms, whichever is appropriate.
 
695
2
Hey, Halls, thanks.
Unless du/dx is a constant we cannot just shift du/dx inside or outside the integral.
Yes, I absolutely agree! The intention was, as you say next, choosing one form or the other, depending on which derivative I find easier to get.
 

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